Question

ABC is a triangle in which $\angle CAB={80}^{\circ }and\angle ABC={50}^{\circ }$. AE, BF and CD are the altitudes and O (intersection point of AE, BF, and CD) is the orthocentre. What is the value of $\angle AOB$?

• A. ${65}^{\circ }$
• B. ${70}^{\circ }$
• C. ${50}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is D ${130}^{\circ }$

$\angle ACB={50}^{\circ }$ $\because \angle A+\angle B+\angle C={180}^{\circ }$
$\angle CFO=\angle CEO={90}^{\circ }$
$\therefore \angle FOE={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{50}^{\circ })={130}^{\circ }$
but $\angle AOB=\angle FOE={130}^{\circ }$